Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $q = \dfrac{-n + 2}{-2n - 10} \div \dfrac{-2n^2 + 4n}{n^2 + 9n + 20} $
Dividing by an expression is the same as multiplying by its inverse. $q = \dfrac{-n + 2}{-2n - 10} \times \dfrac{n^2 + 9n + 20}{-2n^2 + 4n} $ First factor the quadratic. $q = \dfrac{-n + 2}{-2n - 10} \times \dfrac{(n + 5)(n + 4)}{-2n^2 + 4n} $ Then factor out any other terms. $q = \dfrac{-(n - 2)}{-2(n + 5)} \times \dfrac{(n + 5)(n + 4)}{-2n(n - 2)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ -(n - 2) \times (n + 5)(n + 4) } { -2(n + 5) \times -2n(n - 2) } $ $q = \dfrac{ -(n - 2)(n + 5)(n + 4)}{ 4n(n + 5)(n - 2)} $ Notice that $(n - 2)$ and $(n + 5)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -\cancel{(n - 2)}(n + 5)(n + 4)}{ 4n\cancel{(n + 5)}(n - 2)} $ We are dividing by $n + 5$ , so $n + 5 \neq 0$ Therefore, $n \neq -5$ $q = \dfrac{ -\cancel{(n - 2)}\cancel{(n + 5)}(n + 4)}{ 4n\cancel{(n + 5)}\cancel{(n - 2)}} $ We are dividing by $n - 2$ , so $n - 2 \neq 0$ Therefore, $n \neq 2$ $q = \dfrac{-(n + 4)}{4n} ; \space n \neq -5 ; \space n \neq 2 $